Saudi Arabian Mathematical Competitions

Denote by $a, b, c$ the length sides of triangle $ABC$. In triangle $ADM$ we have
$$\begin{gathered} DM^{2} = AM^{2} - AD^{2} = \frac{2\left(b^{2} + c^{2}\right) - a^{2}}{4} - \frac{4K^{2}}{a^{2}} \\ = \frac{2\left(b^{2} + c^{2}\right) - a^{2}}{4} - \frac{16K^{2}}{4a^{2}} \end{gathered}$$
$$\begin{gathered} = \frac{2\left(b^{2} + c^{2}\right) - a^{2}}{4} - \frac{1}{4a^{2}}\left(2 \sum a^{2}b^{2} - \sum a^{4}\right) \\ = \frac{1}{4a^{2}}\left(-2b^{2}c^{2} + b^{4} + c^{4}\right) = \left(\frac{b^{2} - c^{2}}{2a}\right)^{2} \end{gathered}$$
Since $\widehat{B} \geq 2\widehat{C} > \widehat{C}$, it follows $b > c$, hence
$$\begin{equation*} DM = \frac{b^{2} - c^{2}}{2a} \tag{1} \end{equation*}$$
The inequality $DM \geq \frac{AB}{2}$ is equivalent to $\frac{b^{2} - c^{2}}{2a} \geq \frac{c}{2}$, that is $b^{2} \geq c^{2} + ac$. Using the cosine law, the last inequality becomes $a \geq 2c \cos B + c$, or $\sin A \geq 2 \sin C \cos B + \sin C$. We can write $\sin(B + C) \geq 2 \sin C \cos B + \sin C$, hence $\sin(B - C) \geq \sin C$, and we get $2 \sin \frac{B - 2C}{2} \cos \frac{B}{2} \geq 0$. This inequality is true because $\widehat{B} \geq 2\widehat{C}$, and we are done.